Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 - 12z + 27}{-2z^3 + 200z} \div \dfrac{2z - 6}{5z^2 - 50z} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{z^2 - 12z + 27}{-2z^3 + 200z} \times \dfrac{5z^2 - 50z}{2z - 6} $ First factor out any common factors. $q = \dfrac{z^2 - 12z + 27}{-2z(z^2 - 100)} \times \dfrac{5z(z - 10)}{2(z - 3)} $ Then factor the quadratic expressions. $q = \dfrac {(z - 3)(z - 9)} {-2z(z - 10)(z + 10)} \times \dfrac {5z(z - 10)} {2(z - 3)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (z - 3)(z - 9) \times 5z(z - 10)} { -2z(z - 10)(z + 10) \times 2(z - 3)} $ $q = \dfrac {5z(z - 3)(z - 9)(z - 10)} {-4z(z - 10)(z + 10)(z - 3)} $ Notice that $(z - 10)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {5z(z - 3)(z - 9)\cancel{(z - 10)}} {-4z\cancel{(z - 10)}(z + 10)(z - 3)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $q = \dfrac {5z\cancel{(z - 3)}(z - 9)\cancel{(z - 10)}} {-4z\cancel{(z - 10)}(z + 10)\cancel{(z - 3)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $q = \dfrac {5z(z - 9)} {-4z(z + 10)} $ $ q = \dfrac{-5(z - 9)}{4(z + 10)}; z \neq 10; z \neq 3 $